A330915 Sum of the "middle" side lengths (b such that a <= b <= c) of all Heronian triangles with perimeter A051518(n).

4, 5, 5, 8, 12, 23, 45, 15, 29, 13, 48, 30, 77, 24, 69, 117, 25, 25, 46, 119, 20, 26, 110, 246, 26, 167, 172, 205, 169, 79, 468, 33, 229, 38, 222, 167, 429, 41, 429, 101, 270, 560, 416, 100, 153, 276, 390, 717, 50, 615, 61, 61, 60, 404, 634, 214, 130, 130, 1033, 975, 382

Offset: 1

Author

Wesley Ivan Hurt, May 02 2020

Keywords

• nonn

Examples

			a(1) = 4; there is one Heronian triangle with perimeter A051518(1) = 12, which is [3,4,5] and its "middle" side length is 4.
a(6) = 23; there are two Heronian triangles with perimeter A051518(6) = 32, [4,13,15] and [10,10,12]. The sum is 13 + 10 = 23.
		

Links

• Eric Weisstein's World of Mathematics, Heronian Triangle
• Wikipedia, Heronian triangle
• Wikipedia, Integer Triangle

Crossrefs

Cf. A051516, A051518, A070138, A096468, A298079, A298614, A305717.

Cf. A330912, A330916.

Formula

a(n) = Sum_{k=1..floor(c(n)/3)} Sum_{i=k..floor((c(n)-k)/2)} sign(floor((i+k)/(c(n)-i-k+1))) * chi(sqrt((c(n)/2)*(c(n)/2-i)*(c(n)/2-k)*(c(n)/2-(c(n)-i-k)))) * i, where chi(n) = 1 - ceiling(n) + floor(n) and c(n) = A051518(n). - Wesley Ivan Hurt, May 12 2020